Question

Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:
Cu(s) + 2 AgNO3(aq) → Cu (NO3)2(aq) + 2 Ag(s)

a. What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 132.5 grams of AgNO3?

Answer 1

Answer:

ok

Explanation:

percentage yeild of the reaction is 95.12

Answer 2

Answer:

The percent yield of the reaction, if 71.5g of Ag was obtained from 132.5g of AgNO₃, is 84.97%.

Explanation:

Given that,

       Cu + 2AgNO₃   →   Cu(NO₃)₂ + 2Ag

    mass of AgNO₃ = 132.5g

           mass of Ag = 71.5g

We know that,

       atomic mass of Ag = 107.87g/mol

 molar mass of  AgNO₃ = 107.87 + 14.01 + 3×16 = 169.88g/mol.

percentage of Ag in AgNO₃ is given by,

$percent\; of\; Ag \;in\; AgNO_3=\frac{atomic\; mass \;of \;Ag}{molar\;mass \;of \;AgNO_3} \times100$

⇒                                     = (107.87 / 169.88) ×100 = 63.5%

the theoretical yield of Ag from  AgNO₃ is,

               amount of Ag = 132.5 × (63.5/100) = 84.14g

However, the obtained amount of Ag is 71.5g

the percent yield of the reaction is given by,

              $percent \;yield = \frac{actual \;yield}{theoretical\; yield} \times100$

⇒                                 = (71.5 / 84.14) × 100 = 84.97 %

The answer is 84.97%